What are the primes p > 0 for which $1/p$ has a purely periodic decimal expansion with a period$5$ digits long?
[NOTE: $1/37 = 0.027027027027 .....$ starts to repeat immediately, so it is purely periodic. Its period is $3$ digits long.]
Hope u enjoy solving the problem...
For prime lovers
- Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Man himself is the master of his fate...
- Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Re: For prime lovers
Did anyone solve it or make any progress?
Man himself is the master of his fate...
Re: For prime lovers
The period of the repeating decimal of $1/p$ is equal to the order of $10$ modulo $p$. If $10$ is a primitive root modulo $p$, the period is equal to $p − 1$; if not, the period is a factor of $p − 1$. This result can be deduced from Fermat's little theorem, which states that \[10^{p-1}\equiv 1(mod p)\]\
So,all thats left is to find the prime numbers.I hope it was helpful.
So,all thats left is to find the prime numbers.I hope it was helpful.
- Abdul Muntakim Rafi
- Posts:173
- Joined:Tue Mar 29, 2011 10:07 pm
- Location:bangladesh,the earth,milkyway,local group.
Re: For prime lovers
Mahathir, how can you find the answer easily this way?
There is an easier way to find out the primes... You just need a crux move...
Here's it.
There is an easier way to find out the primes... You just need a crux move...
Here's it.
Man himself is the master of his fate...